$$1.00\,g$$  of a non-electrolyte solute ( molar mass $$250\,g\,mo{l^{ - 1}}$$   ) was dissolved in $$51.2\,g$$  of benzene. If the freezing point depression constant, $${k_f}$$  of benzene is $$5.12\,K\,kg\,mo{l^{ - 1}},$$    the freezing point of benzene will be lowered by

Solution :
$$\eqalign{ & {\text{Molality of non - electrolyte solute}} \cr & = \frac{{\frac{{{\text{Weight of solute in }}\left( g \right)}}{{{\text{Molecular weight of solute }}}}}}{{{\text{Weight of solvent in }}(kg)}} \cr & = \frac{{\frac{1}{{250}}}}{{0.0512}} \cr & = \frac{1}{{250 \times 0.0512}} \cr & = 0.0781\,m \cr & \Delta {T_f} = {k_f} \times {\text{molality of solution}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 5.12 \times 0.0781 \approx 0.4K \cr} $$